New Function

It becomes $10^{10} \times 9^{9} \times 8^{8} \times ... \times 2^{2}$

I analysed each term.

$10^{10}$ provides $\boxed{10}$ zeroes.

$9^{9}$ will end in 9, so it provides no zeroes.

$7^{7}$ will end in 3, so it provides no zeroes.

$3^{3}$ will end in 9, so it provides no zeroes.

We have left

$8^{8}$ , which ends in 6

$6^{6}$ , which ends in 6

$5^{5}$

$4^{4}$ , which ends in 6

$2^{2}$ , which is 4

Now, 4 multiplies with two of the fives to give 100, which is $\boxed{2}$ zeroes. We are left with three fives, which multiply with each of the sixes to give 9000, accounting for $\boxed{3}$ zeroes.

Hence the number of zeroes is $10 + 2 + 3 = \boxed{15}$

There are some errors, I'm sorry, this is the first time I'm doing this I also know that this is a very loose proof, but I think I can justify all my steps. I got the last digits by observation of the patterns formed.

To find out the number of zeros, we have to look for the no. of 5's and 2's present in that number.As the number of 2's are greater than the number of 5's. Therefore, we look only for the 5's present in the number. There are 10 5's in $10^{10}$ and 5 5's in $5^{5}$ . Hence, there are total 15 zeros present at the end of the number.

10^10 gives 10 zeros. Then we have 5^5 which can give us five more zeros by multiplying with two "2s" and three "4s" . Rest of the term will not give any zeros ! So the result should be 10+5 = 15